2x-505= -340
x+304=505
x-505=304
720:(x-17)=12
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\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)
\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)
ta co x :2 : 4 + 505 +579
=> x : 2 : 4 = 579 - 505 = 74
=> x:2 = 74 x 4 = 296
=> x = 296m x 2
= 592
Gợi ý: rút gọn cho 101 rồi đặt 5 ra ngoài làm thừa số chung thì sẽ tìm ra kết quả là \(\frac{25}{24}\)
\(\left(\frac{505}{707}+\frac{222}{333}\right)\cdot x=\frac{404}{909}\)
=> \(\left(\frac{5}{7}+\frac{2}{3}\right)\cdot x=\frac{4}{9}\)
=> \(\frac{29}{21}\cdot x=\frac{4}{9}\)
=> \(x=\frac{4}{9}:\frac{29}{21}\)
=> \(x=\frac{28}{87}\)
TK mk nha!
( 505/707+ 222/333) . x = 404/ 909
29/21 . x = 4/9
x= 4/9 : 29/21
x= 4/9 . 21/29
x= 28/87
Vậy x= 28/87
Ta có :\(\frac{230+x}{505+x}=\frac{4}{5}\)
\(\Rightarrow5.\left(230+x\right)=4.\left(505+x\right)\)
\(\Rightarrow1150+5x=2020+4x\)
\(\Rightarrow5x-4x=2020-1150\)
\(\Rightarrow x=870\)
Vậy x =870